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问题描述
Given preorder and inorder traversal of a tree, construct the binary tree.
解决思路
首先确定根节点，然后确定左右子树的节点数目。依次递归即可。
假设输入的序列均合法。
程序
public class BuildTreeFromPreorderAndInorder {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder == null || inorder == null || preorder.length == 0
|| preorder.length != inorder.length) {
return null;
}
int len = preorder.length;
TreeNode root = buildHelper(preorder, 0, len - 1, inorder, 0, len - 1);
return root;
}
private TreeNode buildHelper(int[] preorder, int pre_start, int pre_end,
int[] inorder, int in_start, int in_end) {
// bound
if (pre_start > pre_end || in_start > in_end) {
return null;
}
int rootVal = preorder[pre_start];
TreeNode root = new TreeNode(rootVal);
// find root-val in inorder list
int i = 0;
for (; i <= in_end; i++) {
if (inorder[i] == rootVal) {
break;
}
}
int leftLen = i - in_start;
root.left = buildHelper(preorder, pre_start + 1, pre_start + leftLen,
inorder, in_start, i - 1);
root.right = buildHelper(preorder, pre_start + leftLen + 1, pre_end,
inorder, i + 1, in_end);
return root;
}
}
Follow up
如果给定的是中序和后序遍历序列，如何构建二叉树？
解决思路是类似的。
程序
public class BuildTreeFromInorderAndPostorder {
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (inorder == null || postorder == null || inorder.length == 0
|| inorder.length != postorder.length) {
return null;
}
int len = inorder.length;
TreeNode root = buildHelper(inorder, 0, len - 1, postorder, 0, len-1);
return root;
}
private TreeNode buildHelper(int[] inorder, int in_start, int in_end, int[] postorder,
int post_start, int post_end) {
if (in_start > in_end || post_start > post_end) {
return null;
}
int rootVal = postorder[post_end];
TreeNode root = new TreeNode(rootVal);
int i = in_start;
for (; i <= in_end; i++) {
if (inorder[i] == rootVal) {
break;
}
}
int leftLen = i - in_start;
root.left = buildHelper(inorder, in_start, i-1, postorder, post_start, post_start+leftLen-1);
root.right = buildHelper(inorder, i+1, in_end, postorder, post_start+leftLen, post_end - 1);
return root;
}
}
Construct Binary Tree from Preorder and Inorder Traversal
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原文地址：http://www.cnblogs.com/harrygogo/p/4656412.html
